Problem: Solve for $x$ and $y$ using elimination. ${2x+3y = 20}$ ${-2x-5y = -32}$
We can eliminate $x$ by adding the equations together when the $x$ coefficients have opposite signs. Add the equations together. Notice that the terms $2x$ and $-2x$ cancel out. $-2y = -12$ $\dfrac{-2y}{{-2}} = \dfrac{-12}{{-2}}$ ${y = 6}$ Now that you know ${y = 6}$ , plug it back into $\thinspace {2x+3y = 20}\thinspace$ to find $x$ ${2x + 3}{(6)}{= 20}$ $2x+18 = 20$ $2x+18{-18} = 20{-18}$ $2x = 2$ $\dfrac{2x}{{2}} = \dfrac{2}{{2}}$ ${x = 1}$ You can also plug ${y = 6}$ into $\thinspace {-2x-5y = -32}\thinspace$ and get the same answer for $x$ : ${-2x - 5}{(6)}{= -32}$ ${x = 1}$